The solution of the differential equation lef | vedclass

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(B) Given equation: $\left( {1 + {e^{2y}}} ight){e^{{{	an }^{ - 1}}x}}dx = \left( {1 + {x^2}} ight)\left( {{e^y} + {{\left( {{e^y} - 1} ight)}^

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$y = \ln \left( {	an \left( {y - {e^{{{	an }^{ - 1}}x}} + C} ight)} ight)$

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(B) Given equation: $\left( {1 + {e^{2y}}} ight){e^{{{	an }^{ - 1}}x}}dx = \left( {1 + {x^2}} ight)\left( {{e^y} + {{\left( {{e^y} - 1} ight)}^2}} ight)dy$Rearranging terms: $\frac{{{e^{{{	an }^{ - 1}}x}}}}{{1 + {x^2}}}dx = \frac{{{e^y} + {e^{2y}} - 2{e^y} + 1}}{{1 + {e^{2y}}}}dy = \frac{{{e^{2y}} - {e^y} + 1}}{{1 + {e^{2y}}}}dy$Integrating both sides: $\int \frac{{{e^{{{	an }^{ - 1}}x}}}}{{1 + {x^2}}}dx = \int \frac{{{e^{2y}} - {e^y} + 1}}{{1 + {e^{2y}}}}dy$Let $u = 	an^{-1}x$,then $du = \frac{1}{1+x^2}dx$. The $LHS$ becomes $\int e^u du = e^u = e^{	an^{-1}x}$.For the $RHS$: $\int \frac{e^{2y}+1-e^y}{1+e^{2y}} dy = \int (1 - \frac{e^y}{1+e^{2y}}) dy = y - \int \frac{e^y}{1+(e^y)^2} dy$.Let $v = e^y$,then $dv = e^y dy$. The integral becomes $y - 	an^{-1}(e^y) + C$.Equating both sides: $e^{	an^{-1}x} = y - 	an^{-1}(e^y) + C$.Rearranging: $	an^{-1}(e^y) = y - e^{	an^{-1}x} + C$.Taking tangent on both sides: $e^y = 	an(y - e^{	an^{-1}x} + C)$.Taking natural log: $y = \ln(	an(y - e^{	an^{-1}x} + C))$.

The solution of the differential equation $\left( {1 + {e^{2y}}} \right){e^{{{\tan }^{ - 1}}x}}dx - \left( {1 + {x^2}} \right)\left( {{e^y} + {{\left( {{e^y} - 1} \right)}^2}} \right)dy = 0$ is

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